A cylindrical capacitor is made of two thin-walled concentric cylinders. The inner cylinder has radius 5 mm , and the outer one a radius 11 mm . The common length of the cylinders is 160 m . What is the potential energy stored in this capacitor when a potential difference 6 V is

Answers

Answer 1

Answer:

The  potential energy is [tex]PE = 2.031 *10^{-7} \ J[/tex]

Explanation:

From the question we are told that

    The inner radius is  [tex]r_i = 5 \ mm = 0.005 \ m[/tex]

      The outer radius is  [tex]r_o = 11 \ mm = 0.011 \ m[/tex]

     The  common length is  [tex]l = 160 \ m[/tex]

      The  potential  difference is   [tex]V = 6 \ V[/tex]

Generally the capacitance of the cylindrical capacitor is mathematically represented as

       [tex]C = \frac{2 \pi * k * \epsilon_o }{ ln \frac{ r_o }{r_i} } * l[/tex]

Where  [tex]\epsilon _o[/tex] is the permitivity of free space with the values [tex]\epsilon _o = 8.85*10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]

and  k  is the dielectric constant  of the dielectric material here the  dielectric material is free space so  k  =   1

     Substituting values

             [tex]C = \frac{2* 3.142 * 1 * 8.85*10^{-12} }{ ln \frac{ 0.011}{0.005} } * 160[/tex]

             [tex]C = 1.129 *10^{-8} \ F[/tex]

The potential energy stored is mathematically represented as

       [tex]PE = \frac{1}{2} * C * V ^2[/tex]

substituting values

      [tex]PE = 0.5 * 1.129 *10^{-8} * (6)^2[/tex]

      [tex]PE = 2.031 *10^{-7} \ J[/tex]


Related Questions

What is the voltage output (in V) of a transformer used for rechargeable flashlight batteries, if its primary has 480 turns, its secondary 8 turns, and the input voltage is 118 V

Answers

Answer:

1.97 V

Explanation:

Applying

N1/N2 = V1/V2................... Equation 1

Where N1 = primary turns, N2 = Secondary turns, V1 = primary/input voltage, V2 = secondary/output voltage.

make V2 the subject of the equation

V2 = V1(N2/N1)............. Equation 2

Given: V1 = 118 V, N1 = 480 turns, V2 = 8 turns.

Substitute into equation 2

V2 = 118(8/480)

V2 = 1.97 V.

A source containing a mixture of hydrogen and deuterium atoms emits light at two wavelengths whose mean is 540 nm and whose separation is 0.170 nm. Find the minimum number of lines needed in a diffraction grating that can resolve these lines in the first order.

Answers

Answer:

N=3176.5rulling

Explanation:

We were told that the source containing a mixture of hydrogen and deuterium atoms emits light with

wavelengths whose mean is 540 nm

Then λ= 540 nm, but we need to convert to metre which = (540× 10⁻⁹m)

Also whose separation is 0.170 nm, which mean the difference between the wavelength is 0.170 nm then

Δ λ = 0.170 nm the we convert to metre we have. Δλ= 0.170 nm= (0.170×10⁻⁹m)

the formular below can be used to can be used to calculate our minimum number of lines

N= λ /(m Δλ)

Where N is number of fillings i.e number of lines

λ= wavelength

Δλ= difference in wavelength

m=1

Then if we substitute the values we have

,N= (540× 10⁻⁹ m)/[(1)*(0.170× 10⁻⁹m)]

N =3176.5rulling

Therefore, minimum number of lines = =3176.5rulling

Two large, flat, horizontally oriented plates are parallel to each other, a distance d apart. Half-way between the two plates the electric field field has a magnitude E. If the separation of the plates is reduced to d/2 what is the magnitude of the electric field half-way between the plates

Answers

Answer:

Explanation:

Let the potential difference between the plate is V . Then in the first case

Electric field E between plate

E₁ = V / d

where d is separation between plate

When the plate separation becomes d / 2

Electric field E between plate

E₂ = V / d /2

= 2 V / d =2E₁

Or twice the earlier field

An automotive air conditioner produces a 1-kW cooling effect while consuming 0.75 kW of power. What is the rate at which heat is rejected from this air conditioner

Answers

Answer:

The rejected by the air conditioning system is 1.75 kilowatts.

Explanation:

A air conditioning system is a refrigeration cycle, whose receives heat from cold reservoir with the help of power input before releasing it to hot reservoir. The First Law of Thermodynamics describes the model:

[tex]\dot Q_{L} + \dot W - \dot Q_{H} = 0[/tex]

Where:

[tex]\dot Q_{L}[/tex] - Heat rate from cold reservoir, measured in kilowatts.

[tex]\dot Q_{H}[/tex] - Heat rate liberated to the hot reservoir, measured in kilowatts.

[tex]\dot W[/tex] - Power input, measured in kilowatts.

The heat rejected is now cleared:

[tex]\dot Q_{H} = \dot Q_{L} + \dot W[/tex]

If [tex]\dot Q_{L} = 1\,kW[/tex] and [tex]\dot W = 0.75\,kW[/tex], then:

[tex]\dot Q_{H} = 1\,kW + 0.75\,kW[/tex]

[tex]\dot Q_{H} = 1.75\,kW[/tex]

The rejected by the air conditioning system is 1.75 kilowatts.

Assume that the physics instructor would like to have normal visual acuity from 21 cm out to infinity and that his bifocals rest 2.0 cm from his eye. What is the refractive power of the portion of the lense that will correct the instructors nearsightedness

Answers

This is note the complete question, the complete question is:

One of the lousy things about getting old (prepare yourself!) is that you can be both near-sighted and farsighted at once. Some original defect in the lens of your eye may cause you to only be able to focus on some objects a limited distance away (near-sighted). At the same time, as you age, the lens of your eye becomes more rigid and less able to change its shape. This will stop you from being able to focus on objects that are too close to your eye (far-sighted). Correcting both of these problems at once can be done by using bi-focals, or by placing two lenses in the same set of frames. An old physicist instructor can only focus on objects that lie at distance between 0.47 meters and 5.4 meters.

Assume that the physics instructor would like to have normal visual acuity from 21 cm out to infinity and that his bifocals rest 2.0 cm from his eye. What is the refractive power of the portion of the lense that will correct the instructors nearsightedness?

Answer:  3.04 D

Explanation:

when an object is held 21 cm away from the instructor's eyes, the spectacle lens must produce 0.47m ( the near point) away.

An image of 0.47m from the eye will be ( 47 - 2 )

i.e 45 cm from the spectacle lens since the spectacle lens is 2cm away from the eye.

Also, the image distance will become negative

gap between lense and eye = 2cm

Therefore;

image distance d₁ = - 45cm = - 0.45m

object distance  d₀ = 21 - 2 = 19cm = 0.19m

P = 1/f = 1/ d = 1/d₀ + 1/d₁ = 1/0.19 + (-1/0.45)

P = 1/f =  5.26315789 - 2.22222222

P = 1/f = 3.04093567 ≈ 3.04 D

Each of the boxes starts at rest and is then pulled for 2.0 m across a level, frictionless floor by a rope with the noted force. Which box has the highest final speed

Answers

Answer:

Explanation:

d

An MRI machine needs to detect signals that oscillate at very high frequencies. It does so with an LC circuit containing a 15mH coil. To what value should the capacitance be set to detect a 450 MHz signal?

Answers

Answer:

The capacitance is  [tex]C = 3.2 9 *10^{-16} \ F[/tex]

Explanation:

From the question we are told that

   The  induction of the LC circuit is  [tex]L = 15 mH = 15 *10^{-3} \ H[/tex]

    The  frequency is   [tex]w = 450 \ MHz = 450 *10^{6} \ Hz[/tex]

The natural frequency is mathematically represented as

          [tex]w = \frac{1}{\sqrt{LC} }[/tex]

Where C is the capacitance So  

=>      [tex]C = \frac{1}{L * w^2}[/tex]

substituting values

         [tex]C = \frac{1}{15 *10^{-3} * [450 *10^{6}]^2}[/tex]

         [tex]C = 3.2 9 *10^{-16} \ F[/tex]

The value for which the capacitance should be set to detect a 450 MHz signal is [tex]8.34 \times 10^{-24} \;F[/tex]

Given the following data:

Inductance = 15 mH = [tex]15 \times 10^{-3}\;H[/tex]Frequency = 450 MHz = [tex]450 \times 10^6 \;Hz[/tex]

To determine the value for which the capacitance should be set to detect a 450 MHz signal:

Mathematically, natural frequency is given by the formula:

[tex]f_o = \frac{1}{2\pi \sqrt{LC} }[/tex]

Where:

L is the inductance.C is the capacitance.

Making C the subject of formula, we have:

[tex]C = \frac{1}{(2\pi f_o)^2L} \\\\C = \frac{1}{(2\;\times \;3.142 \times \;450 \;\times\; 10^9)^2 \; \times \;15 \times 10^{-3}}\\\\C = \frac{1}{8 \times 10^{24} \;\times \;15 \times 10^{-3} } \\\\C = \frac{1}{1.2 \times 10^{23}} \\\\C= 8.34 \times 10^{-24} \;F[/tex]

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You are pushing a 60 kg block of ice across the ground. You exert a constant force of 9 N on the block of ice. You let go after pushing it across some distance d, and the block leaves your hand with a velocity of 0.85 m/s. While you are pushing, the work done by friction between the ice and the ground is 3 Nm (3 J). Assuming that the ice block was stationary before you push it, find d.

Answers

Answer: d = 33 cm or 0.33 m

Explanation: In physics, Work is the amount of energy transferred to an object to make it move. It can be expressed by:

W = F.d.cosθ

F is the force applied to the object, d is the displacement and θ is the angle formed between the force and the displacement.

For the ice block, the angle is 0, i.e., force and distance are at the same direction, so:

W = F.d.cos(0)

W = F.d

To determine d:

d = [tex]\frac{W}{F}[/tex]

d = [tex]\frac{3}{9}[/tex]

d = 0.33 m

The distance d the block ice moved is 33 cm.

A 269-turn solenoid is 102 cm long and has a radius of 2.3 cm. It carries a current of 3.9 A. What is the magnetic field inside the solenoid near its center?

Answers

Answer:

Magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T

Explanation:

Given;

number of turns of solenoid, N = 269 turn

length of the solenoid, L = 102 cm = 1.02 m

radius of the solenoid, r = 2.3 cm = 0.023 m

current in the solenoid, I = 3.9 A

Magnitude of the magnetic field inside the solenoid near its centre is calculated as;

[tex]B = \frac{\mu_o NI}{l} \\\\[/tex]

Where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

[tex]B = \frac{4\pi*10^{-7} *269*3.9}{1.02} \\\\B = 1.293 *10^{-3} \ T[/tex]

Therefore, magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T

select the example that best describes a renewable resource.
A.after a shuttle launch, you can smell the jet fuel for hours.
B.solar panels generate electricity that keeps the satellites running.
C.tractor trailers are large trucks that run on diesel fuel.
D. we use our barbeque every night; it cooks with propane.

Answers

Answer:

B.solar panels generate electricity that keeps the satellites running.

Explanation:

Solar panels are a renewable resource because they take energy from the sun.

A proton is accelerated from rest through a potential difference V0 and gains a speed v0. If it were accelerated instead through a potential difference of 2V0, what speed would it gain

Answers

Answer:

Explanation:

Let the charge on proton be q .

energy gain by proton in a field having potential difference of V₀

= V₀ q

Due to gain of energy , its kinetic energy becomes 1/2 m v₀²

where m is mass and v₀ is velocity of proton

V₀ q = 1/2 m v₀²

In the second case , gain of energy in electrical field

= 2 V₀q , if v be the velocity gained in the second case

2 V₀q = 1/2 m v²

1/2 m v² = 2 V₀q = 2 x 1/2 m v₀²

mv² = 2  m v₀²

v = √2 v₀

You indicate that a symbol
is a vector by drawing
A. through the symbol.
B. over the symbol.
c. under the symbol.
D. before the symbol.​

Answers

Answer:

B. over the symbol.

Explanation:

vectors are represented with a symbol carrying an arrow head with also indicates direction

1/1
2. Fluid flows at 2.0 m/s through a pipe of diameter 3.0 cm. What is the
volume flow rate of the fluid in m^3/s *​

Answers

Answer:

9.42*10^-4 m^3/s

Explanation:

d=3/100 m

=0.03 m

A=3.14*0.03^2/6

=4.71*10^-4 m^2

Volume flow rate V = A * s

= 4.71*10^-4 * 2

= 9.42*10^-4

So, the volume flw rate of fluid is 9.42*10^-4 m^3/s

Calculate the maximum kinetic energy of electrons ejected from this surface by electromagnetic radiation of wavelength 236 nm.

Answers

Answer:

Explanation:

Using E= hc/wavelength

6.63 x10^-34 x3x10^8/ 236nm

19.86*10^-26/236*10^-9

=0.08*10^-35Joules

Suppose you have two point charges of opposite sign. As you move them farther and farther apart, the potential energy of this system relative to infinity:_____________.
(a) stays the same.
(b) Increases.
(c) Decreases.
(d) The answer would depend on the path of motion

Answers

Answer:

(b) Increases

Explanation:

The potential energy between two point charges is given as;

[tex]U = F*r = \frac{kq_1q_2}{r}[/tex]

Where;

k is the coulomb's constant

q₁  ans q₂ are the two point charges

r is the distance between the two point charges

Since the two charges have opposite sign;

let q₁ be negative and q₂ be positive

Substitute in these charges we will have

[tex]U = \frac{k(-q_1)(q_2)}{r} \\\\U = - \frac{kq_1q_2}{r}[/tex]

The negative sign in the above equation shows that as the distance between the two charges increases, the potential energy increases as well.

Therefore, as you move the point charges farther and farther apart, the potential energy of this system relative to infinity Increases.

The distance measured between five successive crests of a wave motion executed by a photon of an electromagnetic radiation is 2.4cm.
1. Determine the wavelength in micrometers
2. what is the frequency of the photon
3. what is the energy possessed by the photon.​

Answers

Answer:

1. λ = 0.48 cm = 4800 μm

2. υ = 6.25 x 10¹⁰ Hz

3. E = 4.14 x 10⁻²³ J

Explanation:

1.

Since, the wavelength is defined as the distance between two consecutive or successive crests or troughs. Therefore, in this case the wavelength will be equal to:

Wavelength = λ = Distance between 5 successive crests/5

λ = 2.4 cm/5

λ = 0.48 cm = 4800 μm

2.

The frequency of photon can be given as:

υ = c/λ

where,

υ = frequency of photon = ?

c = speed of light = 3 x 10⁸ m/s

λ = wavelength = 0.48 cm = 0.0048 m

Therefore,

υ = (3 x 10⁸ m/s)/(0.0048 m)

υ = 6.25 x 10¹⁰ Hz

3.

Now, the energy of photon is given as:

E = hυ

where,

E = Energy = ?

h = Plank's Constant = 6.626 x 10⁻³⁴ J.s

Therefore,

E = (6.626 x 10⁻³⁴ J.s)(6.25 x 10¹⁰ Hz)

E = 4.14 x 10⁻²³ J

A 2.0 kg handbag is released from the top of the Leaning Tower of Pisa, and 55 m before reaching the ground, it carries a speed of 29 m / s. What was the average force of air resistance?

Answers

Answer:

4.31 N

Explanation:

Given:

Δy = -55 m

v₀ = 0 m/s

v = -29 m/s

Find: a

v² = v₀² + 2aΔy

(-29 m/s)² = (0 m/s)² + 2a (-55 m)

a = -7.65 m/s²

Sum of forces in the y direction:

∑F = ma

R − mg = ma

R = m (g + a)

R = (2.0 kg) (9.8 m/s² − 7.65 m/s²)

R = 4.31 N

A −3.0 nC charge is on the x-axis at x=−9 cm and a +4.0 nC charge is on the x-axis at x=16 cm. At what point or points on the y-axis is the electric potential zero?

Answers

Answer:

y = 10.2 m

Explanation:

It is given that,

Charge, [tex]q_1=-3\ nC[/tex]

It is placed at a distance of 9 cm at x axis

Charge, [tex]q_2=+4\ nC[/tex]

It is placed at a distance of 16 cm at x axis

We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

[tex]\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0[/tex]

Here,

[tex]r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}[/tex]

So,

[tex]\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}[/tex]

Squaring both sides,

[tex]3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m[/tex]

So, at a distance of 10.2 m on the y axis the electric potential equals 0.

At a distance of 10.2 m, the electric potential equals zero.

According to the question,

Charge,

[tex]q_1 = -3 \ nC[/tex] (9 cm at x-axis)[tex]q_2 = +4 \ nC[/tex] (16 cm at x-axis)

Now,

→ [tex]\frac{kq_1}{r_1} +\frac{kq_2}{r_2} =0[/tex]

or,

→ [tex]\frac{kq_1}{r_1} =-\frac{kq_2}{r_2}[/tex]

→  [tex]\frac{q_1}{r_1} = \frac{q_2}{r_2}[/tex]

here,

[tex]r_1 = \sqrt{y^2+81}[/tex]

[tex]r^2 = \sqrt{y^2+225}[/tex]

By substituting the values,

→      [tex]\frac{-3 }{\sqrt{y^2+225} } = -\frac{4}{\sqrt{y^2+225} }[/tex]

By applying cross-multiplication,

[tex]3\times \sqrt{y^2+225} = 4\times \sqrt{y^2+81}[/tex]

By squaring both sides, we get

→  [tex]9(y^2+225) = 16(y^2+81)[/tex]

    [tex]9y^2+2025 = 16 y^2+1296[/tex]

   [tex]2025-1296=7y^2[/tex]

               [tex]7y^2=729[/tex]

                  [tex]y = 10.2 \ m[/tex]

Thus the solution above is correct.

Learn more about charge here:

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Which of the following statement regarding orbits is true?
A) In a orbit, the satellite and the central body are the two Foci of the eclipse.
B) the orbit of the planet has an elliptical shape with the Sun at one Focus.
C) an elliptical orbit there is one focus and the satellite is located there.
D) the sun and a planet are the two Foci of an orbit

Answers

The correct option is option (B)

The orbit of the planets is elliptical with the sun at one of the foci.

Orbit of planets:

According to Kepler's Laws:

The orbit of a planet around the Sun is an ellipse, with the Sun in one of the focal points of that ellipse. The planet's orbit lies in a plane, called the orbital plane. The point on the orbit closest to the attracting body is the periapsis. The point farthest from the attracting body is called the apoapsis. As the planet moves in its orbit, the line from the Sun to the planet sweeps a constant area of the orbital plane for a given period of time. This means that the planet moves faster near its perihelion than near its aphelion, because at the smaller distance it needs to trace a greater arc to cover the same area. This law is usually stated as equal areas in equal time.For a given orbit, the ratio of the cube of its semi-major axis to the square of its period is constant.

Learn more about Kepler's Laws:

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Balls A and B attract each other gravitationally with a force of magnitude F at distance R. If we triple the mass of ball B and triple the separation of the balls to 3R, what is the magnitude of their attractive force now

Answers

Answer:

F₂ = 1/3 F

Explanation:

Using the law of gravitation of force to solve this question. The law states that the Force of attraction between two masses is directly proportional to the product of their masses and inversely proportional to the square of the distances between them.

Mathematically, F = GMaMb/R² ... 1

G is the gravitational constant

Ma and Mb are the masses of the balls

R is the distance between the balls

If the mass of ball B is tripled and the magnitude of the separation of the balls is increased to 3R, the force between them will be;

F₂ = GMa(3Mb)/(3R)²

F₂ = 3GMaMb/9R² ... 2

Dividing equation 1 by 2 we will have;

F₂/F = (3GMaMb/9R²)/GMaMb/R²

F₂/F =  3GMaMb/9R² * GMaMb/R²

F₂/F = 3/9

F₂/F = 1/3

F₂ = 1/3 F

This shows that the magnitude of the new attractive force is one-third that of the initial attractive force

g An object with mass 1kg travels at 3 m/s and collides with a stationary object whose mass is 0.5kg. The two objects stick together and continue to move. What is the velocity of the two objects together after collision

Answers

Answer:

2

Explanation:

since the second object was in it stationary, it velocity is 0 m/s

A bowling ball of mass 5 kg rolls off the edge of a building 20 meters tall. What is the work done by gravity during the fall, in Joules

Answers

Answer:

1000j

Explanation:

work done = force x distance

w = 5 x 10 x 20 = 1000joules

You are given two infinite, parallel wires, each carrying current.The wires are separated by a distance, and the current in the two wires is flowing in the same direction. This problem concerns the force per unit length between the wires.
A. Is the force between the wires attractive orrepulsive?
B. What is the force per unit length between the two wires?

Answers

Answer:

A. Attractive

B. ( μ₀I² ) / ( 2πd )

Explanation:

A. We know that currents in the same direction attract, and currents in the opposite direction repel, according to ampere's law. In this case the current in the two wires are flowing in the same direction, and hence the force between the two wires are attractive.

B. Suppose that two wires of length [tex]l_1[/tex] and [tex]l_2[/tex] both carry the current [tex]I[/tex] in the same direction ( given ). In the presence of a magnetic field produced by wire 1, a force of magnitude m say, is experienced by wire 2. The magnitude of the magnetic field produced by wire 1 at distance say d, from it's axis, should thus be the following -

[tex]B_1[/tex] = μ₀I / 2πd

The force experienced by wire 2 should thus be -

[tex]F_2[/tex] = I( [tex]l_2[/tex] [tex]*[/tex] [tex]B_1[/tex] )

= I [tex]*[/tex] [tex]l_2 * B_1 *[/tex] Sin( 90 )

= I [tex]*[/tex] [tex]l_2[/tex] ( μ₀I / 2πd )

Therefore the force per unit length experienced by wire 2 toward wire 1 should be ...

( [tex]F_2[/tex] / [tex]l_2[/tex] ) = ( μ₀I² ) / ( 2πd ) ... which is our solution

You have a resistor and a capacitor of unknown values. First, you charge the capacitor and discharge it through the resistor. By monitoring the capacitor voltage on an oscilloscope, you see that the voltage decays to half its initial value in 3.40 msms . You then use the resistor and capacitor to make a low-pass filter. What is the crossover frequency fcfc

Answers

Answer:

The frequency is    [tex]f = 0.221 \ Hz[/tex]

Explanation:

From the question we are told that  

     The  time taken for it to decay to half its original size is [tex]t = 3.40 \ ms = 3.40 *10^{-3} \ s[/tex]

Let the voltage of the capacitor when it is fully charged be  [tex]V_o[/tex]

Then the voltage of the capacitor at time t is  said to be  [tex]V = \frac{V_o}{2}[/tex]

   Now  this voltage can be  mathematical represented as

      [tex]V = V_o * e ^{-\frac{t}{RC} }[/tex]

Where  RC  is the time constant

   substituting values  

    [tex]\frac{V_o}{2} = V_o * e ^{-\frac{3.40 *10^{-3}}{RC} }[/tex]

    [tex]0.5 = e^{-\frac{3.40 *10^{-3}}{RC} }[/tex]

    [tex]- \frac{0.5}{RC} = ln (0.5)[/tex]

     [tex]-\frac{0.5}{RC} = -0.6931[/tex]

     [tex]RC = 0.721[/tex]

Generally the cross-over frequency for a low pass filter is mathematically represented as

          [tex]f = \frac{1}{2 \pi * RC }[/tex]

substituting values  

           [tex]f = \frac{1}{2* 3.142 * 0.72 }[/tex]

           [tex]f = 0.221 \ Hz[/tex]

Waves from two slits are in phase at the slits and travel to a distant screen to produce the second minimum of the interference pattern. The difference in the distance traveled by the wave is:

Answers

Answer:

Three halves of a wavelength I.e 7lambda/2

Explanation:

See attached file pls

Consider the following spectrum where two colorful lines (A and B) are positioned on a dark background. The violet end of the spectrum is on the left and the red end of the spectrum is on the right. A B 5. (1 point) What is the name for this type of spectrum? 6. (1 point) Transition A is associated with an electron moving between the n= 1 and n= 3 levels. Transition B is associated with an electron moving between the n= 2 and n= 5 levels. Which transition is associated with a photon of longer wavelength?

Answers

Answer:

Explanation:

a )

This type of spectrum is called line emission spectrum . Because it consists of lines . It is emission spectrum because it is due to emission of radiation from a source .

b ) The wavelength of a photon  is inversely proportional to its energy .  Photon  due to transition between n = 1 and n = 3 will have higher energy than

that due to transition between n = 2 and n = 5 . So the later photon ( B)  will have greater wavelength or photon  due to transition between n = 2 and n = 5 will have greater wavelength .

Shelly experiences a backward jolt when the driver starts the school bus. Which of the following explains this phenomenon?

Answers

Answer:

A. the inertia of shelly

Explanation:

Answer:

A.  the inertia of Shelly

Explanation:

newtons first law states that an object at rest stays at rest unless on opposite force act against it, so when the buss started to move it acted on shelly forcing her to move back. Shelly's inertia was at a halt and was forced back by the motion of the bus.

EXAMPLE 5 Find the radius of gyration about the x-axis of a homogeneous disk D with density rho(x, y) = rho, center the origin, and radius a. SOLUTION The mass of the disk is m = rhoπa2, so from these equations we have 2 = Ix m = 1 4​πrhoa4 rhoπa2 = a2 4​ .

Answers

Answer:

Radius of gyration = a/2.

Explanation:

So, from the question above I can see that the you are already answering the question and you are stuck up or maybe that's how the problem is set from the start. Do not worry, you are covered in any of the ways. So, from the question we have that;

"The mass of the disk is m = ρπa^2, so from these equations we have y^2 = Ix/m."

(NB: I changed the "rho" word to its symbol).

Thus, the radius of gyration with respect to x-axis = (1/4 πρa^4)/ πρa^2 = a^2/4.

Therefore, the Radius of gyration = a/2.

If a bicycle starts from rest and is pedaled normally until the bike is moving at 6 meters per second across level ground, what kinds of energy have its tires been given? (Select all that apply) g

Answers

Answer: Translational Kinetic Energy

Rotational Kinetic Energy

Explanation:

An object has translational kinetic energy when it is undergoing through a linear displacement.

Rotational energy is kinetic energy due to the rotation of an object .

Here the wheel of bicycle undergoes both translational and rotational kinetic energy has it moves with linear displacement with rotation in it.

Hence, the tires have been two kinds of energy : translational and rotational kinetic energy

if a speed sound in air at o°c is 331m/s. what will be its value at 35 °c​

Answers

Answer:

352 m/s

Explanation:

The velocity of sound in air is approximated as:

v ≈ 331 + 0.6 T

where v is the velocity in m/s and T is the temperature in Celsius.

At T = 35:

v = 331 + 0.6 (35)

v = 352 m/s

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